Coin on a turntable problem in a noninertial frame

[Natchanon]

1. The problem statement
A small coin of mass m is placed on a turntable a distance a from the axis. Imagine that the surface is a little sticky, so the coin does not slide immediately. The turntable (initially at rest) is rotated at constant angular acceleration α, beginning at t=0.
a.) Find the three fictitious forces as functions of time. (find both magnitudes and directions in polar unit vectors)
b.) If you are given the static friction coefficient μ, find the time elapsed before the coin begins to slide. (Hint: This will depend on the net lateral force on the co in.)

Homework Equations

ma_rot = f - mω x (ω x r) - 2mω x v -mα x r

The Attempt at a Solution

I want to know if I understand this correctly. The coin will start to slide when the net force keeping it in circular motion, that is friction + coriolis + centrifugal + Euler, equals zero, correct? Does that mean all I need to do is to find the net force from its x and y components and set it equal to zero?

 

[haruspex]

The coin will start to slide when the net force keeping it in circular motion, that is friction + coriolis + centrifugal + Euler, equals zero, correct?

No, that's wrong in a couple of ways.

If the turntable were stationary, the net force on the coin would be zero, but it won't slide.
What is the equation relating the force of static friction to its coefficient? Be careful, it is more complicated than for kinetic.

Secondly, it is not only the radial forces that need to be considered. If a car brakes on a bend it is more likely to skid - why?

 

[Natchanon]

No, that's wrong in a couple of ways.

If the turntable were stationary, the net force on the coin would be zero, but it won't slide.
What is the equation relating the force of static friction to its coefficient? Be careful, it is more complicated than for kinetic.

Secondly, it is not only the radial forces that need to be considered. If a car brakes on a bend it is more likely to skid - why?

friction force = μmg. If a car brakes on a bend it is more likely to skid because your angular velocity increases in magnitude, so the centrifugal force becomes so large that the friction between the tires and road can't keep you on the road.

 

[haruspex]

friction force = μmg

No, as I wrote, it is more complicated than that.
E.g. what is the frictional force on a block sitting on a level floor with nothing pushing from the side?

If a car brakes on a bend it is more likely to skid because your angular velocity increases in magnitude

No, the angular velocity would start to reduce, yet immediately it wouid be more likely to skid. What has changed in terms of linear acceleration?

 

[Natchanon]

No, as I wrote, it is more complicated than that.
E.g. what is the frictional force on a block sitting on a level floor with nothing pushing from the side?

No, the angular velocity would start to reduce, yet immediately it wouid be more likely to skid. What has changed in terms of linear acceleration?

It increases to the opposite direction of the linear velocity.

 

[haruspex]

It increases to the opposite direction of the linear velocity.

Right, so what happens to the magnitude of the net acceleration?

 

[Natchanon]

Right, so what happens to the magnitude of the net acceleration?

It decreases.

 

[Natchanon]

Right, so what happens to the magnitude of the net acceleration?

I mean "increases".

 

[haruspex]

I mean "increases".

Right, and that increases the demand for static friction if skidding is to be avoided.
This is what the hint was for: the static friction has to supply the net acceleration, which includes both a radial and a tangential component (in the lab frame).
Indeed, you did list the Euler force, so you had that covered, but you referred to the list as the forces "required to keep it in circular motion". Only the radial force is needed for that.

On the other matter, the correct equation for the force of static friction is |F_{static friction}|≤μ_s|N|. Thus, the net of the forces you listed will be zero right up until the point where slipping occurs.

 

[Natchanon]

Right, and that increases the demand for static friction if skidding is to be avoided.
This is what the hint was for: the static friction has to supply the net acceleration, which includes both a radial and a tangential component (in the lab frame).
Indeed, you did list the Euler force, so you had that covered, but you referred to the list as the forces "required to keep it in circular motion". Only the radial force is needed for that.

On the other matter, the correct equation for the force of static friction is |F_{static friction}|≤μ_s|N|. Thus, the net of the forces you listed will be zero right up until the point where slipping occurs.

So that means the coin slides when Euler force + Centrifugal force = Frictional force.

 

[haruspex]

So that means the coin slides when Euler force + Centrifugal force = Frictional force.

No, you are still missing my point about static friction. That equality will hold true at all times that the coin is not sliding.
Also, it should take the form ΣF=0: When on the point of sliding, Euler force + Centrifugal force + Maximum static Frictional force=0.

 

[Natchanon]

No, you are still missing my point about static friction. That equality will hold true at all times that the coin is not sliding.
Also, it should take the form ΣF=0: When on the point of sliding, Euler force + Centrifugal force + Maximum static Frictional force=0.

Oh, I see. So, not frictional force, but the maximum static frictional force which is the threshold. I forgot the minus sign on the right side.

 

[haruspex]

Oh, I see. So, not frictional force, but the maximum static frictional force which is the threshold. I forgot the minus sign on the right side.

Yes.